Although the first three numbers are divisible by ,  doesn't divide evenly into . The next best factor is . The remainder will be . This won't go any further as most of the numbers are even except . Our final answer is just .

We know all of the numbers are divisible by  so when we divide all the numbers by , we have . Next, we can divide al of them by , because the sum of the digits of all numbers are divisible by . So we get . This is as best as we can go so now we multiply the factors to get  as an answer. 

Because they are all even and divisible by , we can divide each number  to get . Next, let's divide by  to get . We are finished as we have a mixture of prime and composite numbers. We multiply the factors to get .

The key here is to find the prime factorizations of both numbers and multiplying the common prime factors together:

Both prime factorizations have in common, so is our answer.

The first seven primes are 2, 3, 5, 7, 11, 13, and 17. Don't forget about 2, the smallest prime number, and also the only even prime! Adding these seven numbers gives a sum of 58, and 58/2 = 29.

There are 8 possible numbers; 4,6,8,10,12,14,16,18.

One is not a prime number, so only 8, 10, 12, 14, 16, and 18 can be the sum of two different prime numbers.

The primes, in order, are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, ...

We create a few series:

 -> series length 2: 2,3

 -> series length 3: 3,5,7

 -> series length 5: 5,7,11,13,17

 -> series length 7: 7,11,13,17,19,23,29

etc.

We can then see that, of the answers, only 47 and 31 remain possibly correct answers.  Now we need to decide which of those two are impossible. 

We could do another series, but the  series has 11 terms requiring us to go further and further up.  If we do this, we'll find that it terminates at 47, meaning that 31 must be the correct answer.

Another way, however, is to notice that 29 is the end of the  series.  Since 31 is the very next prime number, if we start on 11, the series that terminates in 31 would have to have length 7 as well.  Every series after  will thus end on a number larger than 31, meaning we will never finish on a 31.

The value of , or , is the product of  and , so it will be divisible by 1, p, p * p, and nothing else (we know that the p’s are not divisible because they are prime). Therefore p² has exactly three factors.

(Alternatively, we can plug in any prime number for p and see how many factors p² has. For example, if p is 3, then the factors of p², or 9, are 1, 3, and 9.)

Let x represent the smallest of the four numbers. 

Then we can set up the following equation:

Therefore the four numbers are 51, 52, 53, 54.  The only prime in this list is 53.

A prime number is a number with factors of one and itself.

Let's try to find the factors.

It may not be easy to see  as a composite number, but if you know the divisibility rule for  which is double the last digit and subtract from the rest , you will see  is not prime.